Is ${846153}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {846153}= &&{8}\cdot100000+ \\&&{4}\cdot10000+ \\&&{6}\cdot1000+ \\&&{1}\cdot100+ \\&&{5}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {846153}= &&{8}(99999+1)+ \\&&{4}(9999+1)+ \\&&{6}(999+1)+ \\&&{1}(99+1)+ \\&&{5}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {846153}= &&\gray{8\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{1\cdot99}+ \\&&\gray{5\cdot9}+ \\&& {8}+{4}+{6}+{1}+{5}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${846153}$ is divisible by $9$ if ${ 8}+{4}+{6}+{1}+{5}+{3}$ is divisible by $9$ Add the digits of ${846153}$ $ {8}+{4}+{6}+{1}+{5}+{3} = {27} $ If ${27}$ is divisible by $9$ , then ${846153}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${846153}$ must also be divisible by $9$.